Answers

2015-03-09T19:31:51+05:30
X+y=34
xy=128
let the first number=17+h
second number=17-h

product=(17+x)(17-x)
128=289-h²
289-128=h²=168
h=√168
numbers=17+√168=x=29.961481397
y=17-√168=4.038518603

0
is this is a perfect ans
ya
u know it very wel ths is my exam ques
there is a mistake it is not 168 it is 161
ya
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2015-03-09T20:20:42+05:30

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x = 34 - y
so substitute x value in product form:
(34-y) y = 128
34y - y² = 128
y² - 34y + 128 = 0
Applying quadratic equation:

= \frac{-b pm  \sqrt{ b^{2} -4ac} }{2a}

= \frac{-(-34) pm  \sqrt{ -34^{2}-4 * 1* 128 } }{2*1}

= \frac{34 pm \sqrt{1156-512} }{2}

= \frac{34 pm 2 \sqrt{161} }{2}

=17 pm √161

y = 17 + √161 or 17 - √161

x =34-y
=34-17-√161
= 17-√161 or 17 +√161
2 5 2
tysm
cant understand
pm...indicates..+ or -
this is a formula to find the value of x in 2nd degree